How do you find the equation of the line tangent to the graph of #(2x^3 - 1) / (x²)# at the point (1,1)?

Question

How do you find the equation of the line tangent to the graph of #(2x^3 - 1) / (x²)#  at the point (1,1)?

William Abdalla · Accepted Answer

#y = 4x-3# #(2x^3-1)#/#(x^2)#= #2x-1/x^2# the derivative is slope of the tangent and is #2+2/x^3# at x=1 the slope is 4 as point is (1,1) and slope is 4 general equation of a line is #y = mx+c# given #m=4& (x,y)=(1,1)# substituting #1=4+c# #c=-3# the equation of tangent is #y = 4x-3#

Charles Arocha · Answer

undefined To find the equation of the line tangent to the graph of a function at a given point, you need to find the derivative of the function and evaluate it at the given point.

First, find the derivative of the function (2x^3 - 1) / (x²) using the quotient rule. The derivative is given by:

f'(x) = [(2x^3 - 1)(2x²) - (x²)(6x^2)] / (x^4)

Next, evaluate the derivative at the given point (1,1) by substituting x = 1 into the derivative:

f'(1) = [(2(1)^3 - 1)(2(1)^2) - (1^2)(6(1)^2)] / (1^4)

Simplifying this expression gives:

f'(1) = (1 - 1)(2) - (1)(6) / 1

f'(1) = -4

The slope of the tangent line is -4.

Now, use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:

y - y₁ = m(x - x₁)

Substituting the values of the given point (1,1) and the slope (-4) into the equation, we get:

y - 1 = -4(x - 1)

Simplifying this equation gives:

y - 1 = -4x + 4

Finally, rearrange the equation to obtain the slope-intercept form:

y = -4x + 5

Therefore, the equation of the line tangent to the graph of (2x^3 - 1) / (x²) at the point (1,1) is y = -4x + 5.